四点共圆填空一道

如图,在菱形\(ABCD\)中,\(\angle B=60^\circ\),\(AB=4\),点\(E\)在\(BC\)上,\(BE=3CE\),点\(F\)在\(DE\)上,\(\angle AFC=120^\circ\),\(EF>EC\),则\(DF=\)_____.QQ截图20151125170949答案 \(DF=\dfrac 4 7\sqrt{21}\).

方法一QQ截图20151125171222如图,作\(AM\perp BC\)于点\(M\),延长\(AM\),\(DE\),交于点\(N\), \[\triangle MEN \backsim \triangle ADN,\]所以\[AN=\dfrac{8\sqrt 3}{3},\]QQ截图20151125171233由题意可知点\(A\),\(B\),\(C\),\(F\)在同一个\(\odot P\)上, 连接\(AC\),
因为\(\triangle ABC\)为等边三角形,可得\(\odot P\)的直径为\(\dfrac{8\sqrt 3}{3}\).
所以\(AN\)为\(\odot P\)的直径.
在\({\rm Rt}\triangle AND\)和\({\rm Rt}\triangle AFD\)中,\[AD^2=DF\cdot DN,\] 所以\[DF=\dfrac 4 7\sqrt{21}.\]
方法二
作\(AO \perp BC\)于\(O\),以点\(O\)为坐标原点,\(BC\)为\(x\)轴,\(OA\)为\(y\)轴建立直角坐标系. QQ截图20151125171245则\(A(0,2\sqrt 3)\),\(C(2,0)\),\(E(1,0)\),\(D(4,2\sqrt 3)\),\(P(0,\dfrac{2\sqrt 3}{3})\).
直线\(ED\)解析式为:\[y=\dfrac{2\sqrt 3}{3}x-\dfrac{2\sqrt 3}{3}.\]设\(F(a,\dfrac{2\sqrt 3}{3}a-\dfrac{2\sqrt 3}{3})\),由已知得\(PF=PA\),则
\[a^2+(\dfrac{2\sqrt 3}{3}a-\dfrac{2\sqrt 3}{3}-\dfrac{2\sqrt 3}{3})^2=(\dfrac 43\sqrt 3)^2,\] 解得\(a=\dfrac{16}{7}\), 所以\(DF=\dfrac 4 7\sqrt{21}\).    

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